For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.

The Wikipedia page gives examples of proofs along the lines $2=1$ and the primary source appears the book Maxwell, E. A. (1959), Fallacies in mathematics.

the answers to this will turn out to replicate many of the responses to Gowers' famous question on "false beliefs", so I am not so sure if this question should remain open.
–
SuvritApr 22 '12 at 5:46

11

A false proof is not the same as a false belief. One can read a false proof, know for certain that the conclusion is false (so there is no false belief), and still have trouble pinpointing the error.
–
Steven LandsburgApr 22 '12 at 15:36

1

Indeed, a false proof is not the same as a false belief, and at no point did I imply that! But I mentioned Gowers' question, because the top answer's "false proof" (Cayley-Hamilton) also occurred there as one of the answers. (Believing a "false proof" to be true, is a "false belief", and because of that, there is a strong chance of intersection between the two questions) :-)
–
SuvritApr 22 '12 at 22:27

2

I'm surprised no one has mentioned Stallings's false proof of the Poincare Conjecture, in his paper "How Not to Prove the Poincare Conjecture".
–
Steve DApr 30 '12 at 22:13

34 Answers
34

My favorite example is the following proof of the Cayley-Hamilton theorem, which caused me some disconcertion when I was a student. Let $A$ be a square matrix, and call $p(t) = \det(tI - A)$ its characteristic polynomial. Then $p(A) = \det(AI-A) = 0$.

I like this one, invented by T.Clausen in 1827: since $e^{2\pi i n}=1$ for all integers $n$, we have $e^{2\pi i n+1}=e$, which implies $e^{(2\pi i n+1)^2}=(e^{2\pi i n+1})^{2\pi i n+1}=e^{2\pi i n+1}=e$. Now expanding the square at the exponent gives
$$e^{1-4\pi^2n^2+4\pi n i}=e$$
and after simplifying
$$e^{-4\pi^2n^2}=1$$
for all $n$.

In the definition of an equivalence relation $\sim$, the reflexivity of $\sim$ is redundant: Indeed, for any $x$, by the symmetric property we have $x \sim y$ implies $y \sim x$. By transitivity we have $x \sim y$ and $y \sim x$ imply $x \sim x$. Therefore, using only symmetry and transitivity, we obtain reflexivity.

But this proves the result if there is at least one equivalence?
–
David CorwinMar 20 '13 at 18:02

6

As Davidac says you only need that for any $x$ there exists at least one $y$ such that $x \sim y$. I set this as a homework question for my undergraduate groups course every year and the answers systematically ignore the necessary assumption
–
Paul LevyMar 20 '13 at 20:33

Ethan Akin's "proof" that all vector bundles are stably trivial, and hence the $K$-theory of any space must vanish:

Let $V$ be a vector bundle over the base space $B$. Let $T$ be a trivial bundle of the same rank as $V$. To show that $V$ is stably trivial, it suffices to prove that $$V\oplus V=V\oplus T$$.

Let $P$ be the principal bundle associated with $V$. Pull $P$ back over itself to get a bundle $Q$:

Then $Q$ (together with the map to $B$) is the principal bundle associated to $V\oplus V$. But the bundle $Q\rightarrow P$ clearly has a section, namely the diagonal map (viewing $Q$ as a subspace of $P\times P$). Thus $Q=P\times GL_n$, which (together with the same map to $B$) is the principal bundle associated to $V\oplus T$.

I like this. It shows how easy it is to fool yourself and others by drawing a diagram and saying ''the natural map'' and ''canonically isomorphic'' a few times! Apparently, the paper was peer-reviewed, but it states clearly that the purpose was to discuss a fallacious proof.
–
Johannes EbertApr 30 '12 at 19:55

I find the false proof illuminating, since it shows the limitation of a naive treatment of the continuum function $\kappa\mapsto 2^\kappa$. It simply isn't necessarily the case that the two groups have different cardinalities, even though it is necessarily the case that they are not isomorphic.

The theorem is correct, and usually obtained as an application of
contour integration, or of Fourier inversion ($\sin x / x$ is a multiple of
the Fourier transform of the characteristic function of an interval).
The poof, which is the first one I saw
(given in a footnote in an introductory textbook on quantum physics),
is not correct, because the integral does not converge absolutely.
One can rescue it by writing $\int_0^M \sin x \phantom. dx/x$
as a double integral in the same way, obtaining
$$
\int_0^M \sin x \frac{dx}{x} =
\int_0^\infty \frac{dt}{t^2+1}
- \int_0^\infty e^{-Mt} (\cos M + t \cdot \sin M) \frac{dt}{t^2+1}
$$
and showing that the second integral approaches $0$ as $M \rightarrow \infty$;
but this detour makes for a much less appealing alternative to the usual
proof by complex or Fourier analysis.

For more examples and analysis of these "weird fractions", see A Pumping Lemma for Invalid Reductions of Fractions, Michael N. Fried and Mayer Goldberg, The College Mathematics Journal, Vol. 41, No. 5 (November 2010), pp. 357-364.
–
Doug ChathamApr 22 '12 at 21:24

6

My algebra students know better than to fall for this, but they will try to reduce $\frac{x+3}{x+4}$ to $\frac{3}{4}$. So then I invoke this, asking them if $\frac{13}{14}$ reduces to $\frac{3}{4}$, and (when they say No) asking them what happens when $x := 10$.
–
Toby BartelsJun 16 '12 at 15:25

Proof by induction: we prove the statement "All members of any set of people have the same eye color". This is clearly true for any singleton set.

Now, assume we have a set $S$ of people, and the inductive hypothesis is true for all smaller sets. Choose an ordering on the set, and let $S_1$ be the set formed by removing the first person, and $S_2$ be the set formed by removing the last person.

All members of $S_1$ have the same eye color, and also for $S_2$. However, $S_1 \cap S_2$ has members from both sets, so all members of $S$ have the same eyecolor. $\square$

Consider the rational numbers $\mathbb{Q}$ as a totally ordered field. We can add an indeterminate $T_0$ and make it positive but infinitely small (i.e., smaller than positive any element of $\mathbb{Q}$), that is, order $\mathbb{Q}(T_0)$ by lexicographic order of the Laurent series expansion at $0$. Then we can add another indeterminate $T_1$ and make it positive but infinitely small (i.e., smaller than any positive element of $\mathbb{Q}(T_0)$). This process can be iterated transfinitely and we can add $\aleph_1$ indeterminates $T_\iota$ for $\iota<\omega_1$, each infinitely smaller than all the previous ones. The resulting field $K = \mathbb{Q}(T_\iota)$ has cardinality $\aleph_1$ as is easy to show. Now any positive sequence converging to $0$ in $K$ must be eventually constant because it has to cross uncountably many $T_\iota$. So any Cauchy sequence in $K$ is eventually constant. So any Cauchy sequence in $K$ is convergent. So $K$ is complete. But since $K$ contains $\mathbb{Q}$, it contains $\mathbb{R}$. So we have a set of cardinality $\aleph_1$ containing $\mathbb{R}$, which proves the continuum hypothesis.

(The error, of course, is simply that the notion of "completeness" is wrong and its use is nonsense. But if you tell it quickly enough, many people will fall for it.)

One night I proved that every module is flat. Let $M$ be an $R$-module and let $\mathfrak{a}$ be any ideal of the ring $R$. Tensoring the natural inclusion $i:\mathfrak{a} \to R$ we obtain $i_\ast : M \otimes \mathfrak{a} \to M \otimes R$ such that $i_\ast(x\otimes y)=x\otimes i(y)=x\otimes y$, for every $x\in M$ and $y \in \mathfrak{a}$. So $i_\ast$ is injective and we conclude that $M$ is flat...

The heuristic proof use the notion of "raising and lowering subscripts and superscript". Raising subscripts at the left side we obtain
$$b^n=\sum_{k=0}^n\binom{n}{k}a^k=(a+1)^n.$$
Hence, for all $n$,
$$a^n=(b-1)^n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b^k.$$
Lowering exponents, we obtain the inverse relation.

This doesn't look like a false proof. Rather, it's a proof that looks absurd at first glance but that can be made rigorous if you set up the right theoretical framework. Sort of like certain kinds of manipulations with divergent series, or arguments using infinitesimals, or the Dirac delta function.
–
Timothy ChowApr 23 '12 at 14:32

There is a bug with vm_compute and values obtained from functors applications:
using the attached code, I can produce an assumption-free proof of False, or Bus errors.

False proofs in Coq are difficult because Coq produces a "certificate" that can be checked for validity (if one doesn't check the certificate and is happy with the compiler as most people do, it is much easier).

One usual "proof" of Leopoldt Conjecture is that $\mathbb{Z}_p$ is $\mathbb{Z}$-flat, hence the rank of the $p$-adic completion of the units of a number field has the same rank of the units themselves (which is Leopoldt Conjecture) because you can obtain the completion simply as $\mathcal{O}^\times\otimes\mathbb{Z}_p$.

Proof: Let $X$ be a totally disconnected set. If $X$ has only one element, the conclusion clearly follows. Otherwise, for distinct points $a, b \in X$, we have that {$a, b$} $\subset X$ is not connected. Therefore, {$a, b$} admits a separation; but the only way to write this as a disjoint union of nonempty sets is {$a$} $\cup$ {$b$}. Since this gives a separation, each of {$a$} and {$b$} is open. In particular, {$a$} is open for any $a \in X$; so $X$ has the discrete topology. Q.E.D.

Well, the proof would prove more and be much simpler if, instead of looking at the subspace $\{a,b\}$, you just look at the subspace $\{a\}$. Now $\{a\}$ is obviously open, so every topological space whatsoever is discrete.
–
Toby BartelsJun 16 '12 at 15:36

I'm fond of the following false proof of the Strong Law of Large Numbers. Let $X$ be a random variable with expected value $\mu$ and variance $\sigma^2$, and let $X_1, X_2, \dots$ be i.i.d. copies of $X$. Then $$\operatorname{Var} ( \frac{1}{n} \sum_{i=1}^n X_i ) = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n} \rightarrow 0 \textrm{ as } n\rightarrow\infty $$
and since a random variable with variance 0 takes on a single value with probability 1, we must have $$\lim_{n\rightarrow\infty} \frac{1}{n} \sum_{i=1}^n X_i = \mu \textrm{ almost surely.}$$
(It's a memorable heuristic reason to tell undergraduate probability students, even if not a true argument.)

I have known the following for 45 years: in the Euclidian plane, every triangle is isosceles.

The false proof needs a handmade picture; take your pen, it's easy. Start from a triangle $ABC$. Draw the perpendicular bisector of $BC$, and the angle bisector from $A$. Let $I$ be their intersection (if it is not unique, you are done). Let $J$ be the projection of $I$ over $AB$, $K$ that over $AC$. Considering the right triangles $AIJ$ and $AIK$, we see that (lengths) $AJ=AK$, and that $IJ=IK$. Then looking at righttriangles $BIJ$ and $CIK$, we obtain that $BJ=CK$. We conclude that
$$AB=AJ+JB=AK+KC=AC.$$

The falsity is that one of $J$ or $K$ is in the triangle, and the other one is out. Therefore one of the sums above (and only one) should be a difference.

A cavalry sergeant has 24 horses available which he needs to put on 6 carriages. So he needs to divide 24 by 6. He figures that 6 will go into 24 at least once, so he puts down a 1. Subtracting 6 from 24, he gets 18, and he remembers that 18/6=3. So he comes up with the answer 13.

After considerable difficulty with implementing his solution he consults his lieutenant. The lieutenant checks the calculation by evaluating 13*6:

3*6=18
1*6=6

Add them: 24.

Implementation of the result still remains elusive so they consult the colonel, who uses
a different method to check. Write down 13 six times and add.

13

13

13

13

13

13

In adding this up, the colonel arrives at the following sequence of intermediate results:
3,6,9,12,15,18,19,20,21,22,23,24.

$\pi$ is irrational : if $\pi=a/b$ is irreducible, and $a$ is divisible by an odd prime $p$, the series for $\sin \pi =\pi-\pi^3/6+\pi^5/120-\dots$ converges in the $p$-adics, and the limit is obviously not zero, absurd (if $a=2^n$, $n>1$ and the convergence is assured in the 2-adics, with the same contradiction).

True story that I witnessed in a US precalculus class: the teacher told the class that $\pi$ was a rational number, since $\pi = C/d$, where $C$ is the circumference of a circle and $d$ is the diameter. Since $\pi$ can be written as a fraction, it is rational. This still makes me cringe to this day.
–
John EngbersMay 19 '12 at 18:28

Timothy Chow's answer has a nice application. Let $n,x,y,z$ be natural numbers such that $x^n+y^n-z^n=0$. It follows that $e^{x^n+y^n-z^n}=1=e^i$ and the absurd $$1=(e^{x^n+y^n-z^n})^\pi=e^{i\pi}=-1.$$

The Graham Pollak theorem is discussed at this link Combinatorial results without known combinatorial proofs . I came up with a nice short and incomplete proof of it. The tricky part for me was to realize it was incomplete. Follow the commentary if you want to see my "D'oh" moment.
The induction started by taking an a,b complete bipartite subgraph of an (a+b) complete graph.

Some years ago, I came up with this false proof of the irrationality of $\pi$.

It suffices to prove that $x=\pi-3$ is irrational.

For real $y$ with $0\le y\lt1$,
and positive integer $j$, define $d_j(y)$ to be the $j$th digit in the
decimal expansion of $y$.

Let $r_1,r_2,\dots$ be
an enumeration of the rationals in $[0,1)$. The $\it value$ of this
enumeration is $n$ if $d_n(r_n)=d_n(x)$ and $d_j(r_j)\ne d_j(x)$ for $j\lt n$.
If there is no such $n$, then the value of the enumeration is infinite. Note
that if there is an enumeration of infinite value, then $x$ is irrational; it
cannot equal any of the enumerated rationals, as it differs from the first
rational in (at least) the first decimal place, from the second in the second,
etc.

Note also that there are enumerations of arbitrarily large value. For, given
any $n$, you can find $n$ rationals such that the first differs from $x$ in
the first decimal, the second differs from $x$ in the second decimal, and so
on, and then any enumeration that starts off with these $n$ rationals will
have value greater than $n$.

Now, the set of all enumerations of the rationals can be partially ordered by
value; if $E_1$ and $E_2$ are enumerations, then $E_1>E_2$ if the value of
$E_1$ exceeds the value of $E_2$. By Zorn's Lemma, there is an enumeration
maximal with respect to this order. This maximal enumeration cannot have a finite
value --- as we have seen, there are enumerations of arbitrarily great finite
value. So, it must have infinite value. So, $x$ is irrational.

An alternative use for this argument is to apply it to prove that $1/3$ is irrational, the contradiction with the known rationality of $1/3$ thereby establishing that Zorn's Lemma is false.